Talk:Passing to the quotient (function)/Induced is surjective iff function is surjective

Reminder

$\xymatrix{ X \ar[r]^f \ar[d]_w & Y \\ W \ar@{.>}[ur]_{\bar{f} } }$
Diagram

If $f:X\rightarrow Y$ is a function and $w:X\rightarrow W$ another function such that:

• $\forall a,b\in X\left[\big(w(a)=w(b)\big)\implies\big(f(a)=f(b)\big)\right]$

then we may "factor $f$ through $w$ to yield:

• A function, $\bar{f}:W\rightarrow Y$ well defined by $\bar{f}:w\mapsto f(w^{-1}(w))$ such that:
  • $f=\bar{f}\circ w$
• Furthermore, if $w:X\rightarrow W$ is surjective then the induced $\bar{f}:W\rightarrow Y$ is unique].

The $w$ and $W$ represent "whatever", anything.

Claim

• If $f$ is surjective then $\bar{f}$ is surjective

Utility

• Useful property to have when factoring, isn't difficult, can be applied to many things (especially anything involving isomorphisms)
• Motivated by the first group isomorphism theorem, in fact I'm writing this for use there.

Proof

Suppose $f$ is surjective. We wish to show that $\bar{f}$ is too.

• Note: $f$ is surjective means:
  • $\forall y\in Y\exists x\in X[f(x)=y]$

We may now begin the proof. We wish to show $\forall y\in Y\exists a\in W[\bar{f}(a)=y]$

• Let $y\in Y$ be given.
  • By hypothesis: $\exists x\in X[f(x)=y]$
  • Pick $x\in X$ such that $f(x)=y$.
    • Note that by the other part of the hypothesis:
      • $f=\bar{f}\circ w$ so
    • $y=f(x)=(\bar{f}\circ w)(x)=\bar{f}(w(x))$
  • Choose $a\in W$ by $a:=w(x)$
    • Now $\bar{f}(a)=y$
• This completes the proof.

Notes

On paper I've been rather unhappy with this proof, I actually tried to go by contradiction, starting with:

• $\exists y\in Y\forall a\in W[\bar{f}(a)\ne y]$

And hoping to reach a contradiction, but I didn't really "contradict". Obviously I got to "but we see $\bar{f}$ is the negation of (not surjective), but it didn't seem to have a contradiction in it as such.

I'm not sure what I was unhappy about though!